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## Homework Statement

A textbook shows an example of work. Either my understanding of Work and how to calculate it is wrong, or the text is wrong, I am going nutty here and not sure which is wrong, I think it is the text but, then again, I am biased.

The example goes as follows: (between the -----)

(I also included a photo of the example as an attachment)

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A rollerskater of 45kg coasts down a frictionless hill. The hill has an angle of 10 degrees to the horizontal. The horizontal distance from top to bottom is 40m.

A Free Body Diagram Shows Fg vertically downwards and Fgh, perpendicular to Fg, as a horizontal vector. Fgh is here defined as "the component of force of gravity acting in the direction of motion."

Calculations follow under the diagram:

Theta = 10 degrees, Δd = 40m, g=9.8N/kg, m=45kg

W=Fgh x Δd

W=Fg Cos(Theta) x Δd

W=mg Cos(Theta) x Δd

W=45kg x 9.8N/kg x Cos10 x 40m

W=1.7x10^4 J

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## Homework Equations

W=FΔd

## The Attempt at a Solution

I would address the problem like this:

The only two forces acting are(for this problem, no friction or air)

Fg is force of gravity, downwards

Fn is normal force, perpendicular to and upwards from hill surface

Split Fg into two components:

Fgperp is the component of Fg that is perpendicular to the hill surface

Fgpar is the component of Fg that is parallel to the hill surface and downhill

|Fgperp|=|Fn| and they are in opposite directions so they cancel out.

This leaves Fgpar as the only remaining contributor to the net force.

Finding Fgpar:

Fgpar=Fg x Sin(10)

Fgpar=(45kg)(9.8N/kg) x Sin(10)

Fgpar=76.56N

Finding Displacement:

The question clearly says "inclination of the hill is 10 degrees and the horizontal distance from top to bottom is 40m"

So we can say that the hill surface is the hypotenuse of a right-angled triangle with theta=10 and adjacent or x length is 40m. Δd is the hill surface and hypotenuse side.

Since:

Cos(theta)=x/r

rCos(theta)=x

r=x/Cos(theta)

Therefore:

Δd=40m/Cos(10)

Δd=40.62m

Now to find the work done:

Fgpar is in the direction of the displacement.

W=FΔd

W=Fgpar x Δd

W=76.56N x 40.62m

W=3109.87Nm

Is my solution right and the book's wrong? Am I totally missing the boat somehow? Please detail to me what I have misunderstood, if anything.

Thanks to anybody who answers, and please address the conceptual side, I have no difficulty with the calculation parts (apart from thousands of careless mistakes).

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